RESULTS: Erdős #126, graph/Ramsey arm (support run `d0f0aefb`)

Hunt, not a result. Nothing here is evidence for or against RH (docs/08).

Reproduce with .venv/bin/python hunts/support_d0f0aefb/probe.py (a few seconds, stdlib only, writes results.json; re-run on landing and reproduced byte for byte). Every claim marked measured below is produced by that script; every claim marked proved is proved in this file and, where it is checkable on finite data, confirmed by the script.

Notation, following hunts/r_186989/RESULTS.md: $S$ is a set of $k$ primes, $A$ a finite set of distinct positive integers with every off-diagonal sum $a+b$ $S$-smooth, $n = |A|$, and $g(k)$ the maximum $n$. For $p \in S$ let $G_p$ be the graph on $A$ with $a \sim b \iff p \mid a+b$. The briefed premise is that $\bigcup_{p \in S} G_p = K_n$, which is true and is the point of contention.

Landing note (2026-10-21). Landed on main as Hunt #98 from a local-only branch, support/d0f0aefb, that was never pushed; PR #144 counted this arm among the seven that "produced no branch at all". The run's files are as it wrote them apart from punctuation and the dated notes marked as added on landing.

Grading (added on landing, 2026-10-21). Theorems 1 to 4 and Proposition P, labelled proved, are this lab's ordinary arguments, unreviewed and pending external verification; none is kernel-checked. Measured claims are exact checks on the seven r_186989 witnesses and searches in bounded boxes (the coprime table of §3 gives lower bounds only). The two literature bounds behind Theorem 4's door were checked by web search, not read in the sources (the run's HANDBACK.json says so), and the reduction's prior art is unsearched; the run's own advice is to assume it is folklore. In the terms of ALIGNMENT.md §5 the dispositions are a restricted-class obstruction for incidence and cover data (Theorem 3: the cover is realizable at every $n$ with $k=2$ and $\gcd 1$, so no bound on $n$ follows from which pairs are related) and a refuted candidate (r_186989's loose thread 3, §1). Theorem 2 is a bound under a hypothesis and Theorem 4 a one-way reduction; neither improves $3\cdot2^{k-1}$ unconditionally. Two statements did not survive landing and are corrected in place rather than deleted: the level-set clause of Theorem 1, which fails above level 0 and takes the chromatic corollary and the unconditional peel of §3 with it (§2), and §5's claim that a two-variable reduction must enlarge $S$.

Overlap with arms already on main. Theorem 2 is the bound of Hunt #101's primitive descent, $g^*(k)\le 2^k$ (support_7ddfee4b), under the same hypothesis (no element divisible by a prime of $S$), and of Hunt #94's Lemma 2 (support_d176e366, landed with this arm). Theorem 4 belongs next to the four-element reductions of Hunts #93, #95 and #105 (support_8ea74995, support_baf4cde6, support_eccd5f5e), which send each 4-subset of $A$ to a nondegenerate three-term solution without injectivity, a gap all three name; fixing three anchors and letting the fourth element vary makes the map injective, and that is what this arm adds. It also sits beside the two-term reduction $g(k)\le 2+N_2(k)$ of Hunts #96 and #104 (and #97, #99 and #106, landed with this arm), which keeps $S$ fixed too, through the uniformity of Evertse's bound in the coefficients. The refutation of r_186989's loose thread 3 (§1) was on main before this arm landed (Hunts #93, #95, #101, #103), with other witnesses. Theorem 3's point, that the local data is realizable at every size, is the cover-level relative of Hunt #95's Theorem B.

Verdict, in three lines

1. The cover premise is true and it is worthless on its own. There are $k=2$ configurations satisfying "every edge of $K_n$ is covered by some $G_p$", with $\gcd = 1$, for every $n$. So no theorem of the form "cover by $k$ residue graphs $\Rightarrow n \le h(k)$" exists. That is the realizable obstruction the arm asked for, and it is proved, not observed.
2. The graph structure does buy one sharper cover theorem, under a hypothesis: if no element of $A$ shares a factor with $\prod S$, then $n \le 2^k$, which beats Erdős–Turán's $3\cdot 2^{k-1}$ by a factor $3/2$ in that case. Unconditionally the same argument degrades to $n \le 2^k \prod_p T_p$ with $T_p$ the number of distinct values of $v_p$ on $A$, and $T_p$ is not bounded, so it is weaker than Erdős–Turán in general. (Corrected on landing, 2026-10-21: the unconditional form is not established; its proof uses the level-set clause of Theorem 1, which fails above level 0. See the correction after Theorem 1 in §2.)
3. The exact-valuation refinement is not a graph problem at all. Pushed to its end it gives a rigorous reduction, $g(k) \le 3 + N_3(k)$, where $N_3(k)$ counts nondegenerate solutions of the three-term $S$-unit equation $x+y+z=1$ over the same $k$ primes. Erdős #126 follows from $N_3(k) = \exp(o(k))$, and the known lower bounds for such counts are $\exp(k^{1/2+o(1)})$, so nothing known closes this door. It is the same exponential-versus-subexponential gap that Evertse's bound leaves open.

Generic graph methods stall because of (1): the combinatorial shadow of the problem has no bound in it. The content lives in the arithmetic that the shadow throws away.

1. Audit of hunts/r_186989/RESULTS.md

Audited rather than trusted, as instructed. Everything I could check independently holds.

The refuted thread. r_186989 conjectured that the parity argument generalises: $p \notin S$ should cap $|A|$ at $p-1$ by pigeonhole on residues mod $p$ against the pairing $r \leftrightarrow -r$. That pairing argument does not close, because several elements may share one residue class. Explicit counterexample, verified: $S = \{2,5\}$, $p = 3 \notin S$,

$$A = \{1, 4, 124\}: \quad 1+4 = 5, \quad 1+124 = 125 = 5^3, \quad 4+124 = 128 = 2^7 .$$

All three elements are $\equiv 1 \pmod 3$, so no sum is divisible by $3$, and $|A| = 3 > p - 1 = 2$. The correct general statement is much weaker: if $p \notin S$ then the residues of $A$ mod $p$ avoid $0$ and use at most one of each pair $\{r, -r\}$, which bounds nothing, because a single class may hold arbitrarily many elements. The only $p$ for which the argument does bound anything is $p=2$, where every class is its own negative:

Proposition P (proved). If $2 \notin S$ then $|A| \le 2$. Proof. Among any three integers two share a parity, and their sum is even, hence divisible by $2 \notin S$. $\square$

Brute force over $S = \{3,5,7\}$ inside $[1,400)$ finds pairs and no triples, consistent (parity_bound).

2. What $G_p$ actually is (proved, and confirmed on all seven witnesses)

Theorem 1 (structure). Fix $p \in S$ and write $t(a) = v_p(a)$, $a = p^{t(a)}a'$ with $p \nmid a'$. Then, in $G_p$:

• $a \sim b$ for every pair with $t(a) \ne t(b)$, $\min(t(a),t(b)) \ge 1$;
• no pair with $\min(t(a),t(b)) = 0 < \max(t(a),t(b))$ is an edge;
• within a level set $V_t = \{a : t(a) = t\}$, $a \sim b \iff a' + b' \equiv 0 \pmod p$.

Consequently $G_p$ is the disjoint union of $G_p|_{V_0}$ and a complete multipartite join of the graphs $G_p|_{V_t}$, $t \ge 1$; and for odd $p$ each $G_p|_{V_t}$ is a disjoint union of complete bipartite graphs (between the classes $r$ and $-r$, which are distinct since $p \nmid a'$ and $p$ is odd). For $p = 2$ each $G_2|_{V_t}$ is a clique, because the only unit class mod $2$ is its own negative.

Proof. $v_p(a+b) = \min(t(a),t(b))$ when the valuations differ, and $a + b = p^t(a'+b')$ when they agree. $\square$

Correction on landing (2026-10-21). The third clause holds only at level $t=0$. For $t \ge 1$ and $a, b \in V_t$, $a+b = p^t(a'+b')$ is divisible by $p$ whatever $a'+b'$ is, so in $G_p$ every pair of multiples of $p$ is an edge: the multiples of $p$ form one clique, and $G_p$ is the disjoint union of $G_p|_{V_0}$ (complete bipartite blocks between the classes $r$ and $-r$ for odd $p$, a clique for $p=2$) and that clique. The clause as written describes the finer relation $p^{t+1} \mid a+b$, not $G_p$. It fails on the run's own $k=7$ witness: $3, 21 \in V_1$ for $p=3$ have $a'+b' = 1+7 \not\equiv 0 \pmod 3$, yet $3+21 = 24$, and the same happens for $(6,15)$ at $p=3$, $(5,15)$ at $p=5$ and $(7,21)$ at $p=7$. The probe's structure_ok could not see this, because it checks only that each component is a clique, complete bipartite or isolated, which the corrected structure also satisfies. What depends on the clause: the chromatic bound $\chi(G_p) \le 2T_p$ for odd $p$ below (false in general: $A=\{3,15,21,33\}$, $S=\{2,3\}$ is admissible, $G_3 = K_4$ and $T_3 = 1$), and the unconditional peel $n \le \prod_p 2T_p$ of §3, which is therefore not established here. What does not: Theorem 2, whose hypothesis puts every element in $V_0$, and Theorems 3 and 4. The run's measured $\chi \le 2T_p$ on all 27 odd-prime instances stands as a measurement on those gcd-1 witnesses.

Equivalently and more suggestively: mod $p$ the matrix $(a_i + a_j)$ has rank $\le 2$, and $G_p$ is its zero pattern. The complete-bipartite blocks are forced by that rank; there is no freedom in them. This is why the arm's "these are not arbitrary graphs" instinct is right and also why it does not pay: the graphs are so rigid that they carry almost no information.

Measured: the component classification predicted by Theorem 1 holds for every prime of every one of the seven r_186989 witnesses (per_prime.*.structure_ok, no component of any other kind), and $\chi(G_2) = \max(\#\text{odd}, \#\text{even})$ in every case.

The chromatic corollary, and why it dies. Proper colourings of the $G_p$ multiply into a proper colouring of $K_n$, so $n \le \prod_{p\in S}\chi(G_p)$. Theorem 1 gives $\chi(G_p) \le 2T_p$ for odd $p$ (two colours per level set, level sets joined completely; corrected on landing: this step uses the clause that fails above level 0, and the bound is false in general, see the correction after Theorem 1), where $T_p = \#\{v_p(a) : a \in A\}$, and $\chi(G_2) = \max(\#\text{odd},\#\text{even})$. The $p=2$ term alone is $\ge n/2$, so the bound is vacuous on every real witness: measured $\prod_p \chi(G_p)$ is $2, 8, 20, 40, 112, 288, 1920$ against $n = 2,4,5,6,8,10,11$ (product_of_chi). Confirmed for odd $p$: $\chi \le 2T_p$ in all 27 odd-prime instances.

3. The sharper cover theorem it does buy

Theorem 2 (proved). If every $a \in A$ is coprime to $\prod_{p\in S} p$, then $|A| \le 2^k$.

Proof. Induction on $k$. For $k=1$: if $S = \{p\}$ with $p$ odd, $G_p$ is a disjoint union of complete bipartite graphs by Theorem 1 ($T_p = 1$), so its vertex set splits into two independent sets $B_1, B_2$; a pair inside $B_i$ has a sum with empty prime support, which is impossible for distinct positives, so $|B_i| \le 1$ and $|A| \le 2$. If $S = \{2\}$, all elements are odd and $|A| \le 2$ is r_186989's $g(1) = 2$ argument. For $k > 1$ pick an odd $p \in S$ if one exists; $G_p$ splits $A$ into two $G_p$-independent sets, each of whose pairwise sums is $(S\setminus\{p\})$-smooth and whose elements are still coprime to the rest of $S$, so each has size $\le 2^{k-1}$. If $S$ has no odd prime then $k=1$. $\square$

$2^k < 3\cdot2^{k-1}$, so this is a genuine sharpening of the 1934 bound, on the sub-family where it applies. Measured tightness (exhaustive branch and bound over $[1,N]$, elements restricted to be coprime to $\prod S$):

Tight at $k \le 2$, slack from $k=3$ on. (Lower bounds only: bounded universe.)

Why the hypothesis cannot be normalised away, which is the honest limit of this route. Unconditionally the same peel gives $n \le \big(\prod_{p} 2T_p\big)$ (corrected on landing, 2026-10-21: not established, since the peel above level 0 needs the bipartite level-set structure that the correction after Theorem 1 withdraws; the multiples of $p$ form a single clique in $G_p$), and $T_p$ is unbounded: $\{1,2,4\}$ with $S = \{2,3,5\}$ has $T_2 = 3$, and the r_186989 witnesses themselves reach $T_3 = 3$ ($k=6$) and $T_7 = 3$ ($k=7$). The obstruction is exact and structural: if $A_0 = \{a \in A : p \mid a\}$ then $A_0/p$ is again a set with all pairwise sums $S$-smooth over the same $S$, so the level structure is self-similar and no induction on $k$ sees it. Dividing by $\gcd(A)$ (legitimate: $\gcd(A)$ divides an $S$-smooth sum, hence is $S$-smooth) removes only the global case.

4. The realizable obstruction: the cover premise has no bound in it

Theorem 3 (proved). For every $n$ there is a set $A$ of $n$ distinct positive integers with $\gcd(A) = 1$ and a set $S$ of $k=2$ primes such that $\bigcup_{p\in S} G_p = K_n$.

Proof. $A = \{1\} \cup \{2(3j+1) : 0 \le j \le n-2\}$, $S = \{2,3\}$. Any two elements of the second block are even, so $2$ covers that edge. An edge from $1$ to $2(3j+1)$ has sum $6j + 3$, so $3$ covers it. $\gcd = 1$. $\square$

Verified for $n \in \{5,20,60\}$ (residue_cover_unbounded). These sets are of course not $S$-smooth-sum sets (actually_good: false), and that is exactly the point:

The map "configuration $\mapsto$ its cover data" is not injective enough to support any bound. Ramsey, container, rank and cover arguments all take as input which pairs are related, and this input is realizable at every $n$ with $k=2$. Any argument that bounds $n$ must therefore use information that survives past the cover: the exact valuations, or equivalently the sums themselves. The same is true of the weaker relaxation "each $G_p$ is a residue graph mod $p$", which is what Theorem 1 supplies, since the family above is built from residue graphs.

A second, softer stall, measured on real witnesses rather than constructed: the class-$0$ cliques make $\chi(G_p)$ as large as $n-1$ even for genuine optimal witnesses ($k=4$, $A = \{1,2,3,5,7,13\}$: $G_2$ is a clique on the five odd elements). So the one general graph inequality available here, the chromatic product, is not merely weak but vacuous.

5. Where the exact valuation actually leads: a reduction, not a graph

The valuation data is far stronger than the graph data: knowing every $v_p(a_i+a_j)$ determines every sum, and three sums determine three elements ($2a_1 = (a_1+a_2)+(a_1+a_3)-(a_2+a_3)$). So the configuration has $O(1)$ degrees of freedom once a few sums are fixed, which is the opposite of the regime where extremal graph theory has anything to say. Following that to its end:

Theorem 4 (proved). Let $N_3(k)$ be the maximum, over sets $S$ of $k$ primes, of the number of solutions $(x,y,z)$ of $x+y+z=1$ in $S$-units of $\mathbb{Q}$ with no vanishing subsum. Then $$g(k) \;\le\; 3 + N_3(k).$$

Proof. Let $A$ be an $S$-smooth-sum set, $n \ge 4$, and fix three elements $a_1,a_2,a_3$. Put $c = a_2+a_3$ and $c' = a_1+a_3$, both $S$-smooth. For any other $a_i$ put $s_i = a_1+a_i$ and $t_i = a_2+a_i$, both $S$-smooth. Since $s_i - t_i = a_1 - a_2 = c' - c$, we get $s_i + c = t_i + c'$, and dividing by $t_i$, $$X + Y + Z = 1, \qquad X = \frac{s_i}{t_i},\quad Y = \frac{c}{t_i},\quad Z = -\frac{c'}{t_i},$$ a ratio of $S$-smooth integers in each coordinate, hence $S$-units. No subsum vanishes: $X+Y = (t_i+c')/t_i > 0$, $X+Z = (a_i-a_3)/t_i \ne 0$, and $Y+Z = (a_2-a_1)/t_i \ne 0$. The map $a_i \mapsto (X,Y,Z)$ is injective because $Y$ determines $t_i$ and $t_i$ determines $a_i$. $\square$

Measured: the triples are constructed, checked to be $S$-units, checked nondegenerate and checked pairwise distinct for all six applicable witnesses (s_unit_reduction[*].all_checks_pass, counts $n-3$ as claimed). Example, $k=3$, $A = \{1,3,7,17,47\}$: $(9/10,\,1/2,\,-2/5)$ and $(24/25,\,1/5,\,-4/25)$.

Check the implication direction, since the brief asks for it explicitly. Theorem 4 runs one way only. $N_3(k) = \exp(o(k))$ implies Erdős #126; Erdős #126 does not imply anything about $N_3$, since a solution triple need not come from a set $A$. This is the mirror image of r_186989's arm 3, which found a proposed lemma whose truth would refute the conjecture. Here the proposed lemma would prove it.

Is the door open? As far as the literature we could check goes, yes.

• The three-variable trick is what keeps $S$ fixed. The naive two-variable reduction $s_i - t_i = a_1 - a_2$ needs the primes of $a_1-a_2$ adjoined to $S$, and $\omega(a_1-a_2)$ is unbounded, so it proves nothing. Using $c' - c$ for the same difference makes every quantity in the equation $S$-smooth. That is the whole content of the third term. (Corrected on landing, 2026-10-21: the two-variable reduction does not need $S$ enlarged. Evertse's 1984 bound for $\alpha X + \beta Y = 1$ is uniform in the coefficients $\alpha, \beta \in \mathbb{Q}^$, so $s_i - t_i = a_1 - a_2$ can keep $a_1-a_2$ in the coefficient at rank $k$; that is the reduction $g(k) \le 2 + N_2(k)$ of Hunts #96 and #104, and of #97, #99 and #106 landed with this arm. Even a coefficient-free version adjoins $a_1 - a_2$ as one generator, not its $\omega(a_1-a_2)$ primes (Hunt #104's rank $k+1$). Theorem 4 stands; what the third term buys is a coefficient-free equation over the same $S$, not the fixing of $S$.)*
• The best general upper bound is exponential in $k$: Evertse's $3\cdot 7^{2s+3}$ for the two-term equation, and the Evertse–Schlickewei–Schmidt subspace bounds for more terms. Substituted into Theorem 4 these give a bound far worse than $3\cdot2^{k-1}$, so the reduction does not currently improve anything numerically.
• The best known lower bounds are subexponential: Erdős–Stewart–Tijdeman exhibit $S$ with $\exp\{(4+o(1))(s/\log s)^{1/2}\}$ solutions of $ax+by=1$. Since $x+y=1 \Rightarrow x + y/2 + y/2 = 1$ is nondegenerate and injective when $2 \in S$, $N_3(k) \ge N_2(k)$, so the same lower bound applies to $N_3$. It is $\exp(o(k))$, hence consistent with the conjecture.

So Erdős #126 sits inside the same open gap as $S$-unit counting: upper bounds exponential in $k$, lower bounds $\exp(k^{1/2+o(1)})$, ninety years of nobody closing it. That is a sharper statement of the wall than "the $S$-unit machinery is a different information class", which is where r_186989's door list stopped.

Novelty is unsearched. Theorem 4 is derived here and is original to this run in the provenance sense, but three-term $S$-unit reductions of this shape are standard technique and we would expect this to be folklore in the Győry–Evertse literature. Do not claim it as new without a literature search.

6. What I could not settle

• No unconditional improvement on $3\cdot2^{k-1}$. Theorem 2 needs coprimality to $\prod S$, and §3 explains precisely why the hypothesis resists normalisation: the $p$-divisible part is a scaled copy of the whole problem.
• No Ramsey or container bound, and §4 says there cannot be one at the level of the cover. I did not investigate whether a container argument applied to the sums rather than the pairs could work; that is a different object.
• $N_3(k)$'s true growth. Not settled and not cheaply settleable. A finite count of small-height solutions would not bound it.
• Whether Theorem 4 is known. No literature search was run beyond confirming the two cited bounds.

The doors

This arm measured a ceiling (the cover-theoretic ceiling), so it owes the list.

1. Active constraints at the optimum.

2. Frozen-constant inventory.

3. Information class. Everything in §§2–4 stays inside the cover data and is therefore under the ceiling Theorem 3 sets: it cannot produce a bound. Theorem 2 escapes only by adding an arithmetic hypothesis. Theorem 4 changes information class deliberately, reading the sums rather than the incidences, and that is the only door here with a path to the conjecture. The top-ranked door is the three-term $S$-unit count $N_3(k)$, and specifically whether the positivity and fixed-coefficient structure that Theorem 4's solutions carry allows a subexponential count where the general equation does not.

Loose threads

• Bound $T_p$ in terms of $k$. Each extra $p$-adic level in $A$ appears to cost new primes ($1+p^j$ must be $S$-smooth for the powers-of-$p$ family). If $T_p \le C$ absolutely, or even $T_p \le k^{o(1)}$, Theorem 2 becomes unconditional and near-$2^k$. (Landing note: this rests on the unconditional peel of §3, which the correction after Theorem 1 leaves unestablished; bounding $T_p$ alone would not suffice as stated.) First step: measure $\max_p T_p$ over exhaustive optimal witnesses for $k \le 6$, then try to prove $T_p \le 1 + \log_p(\text{something in } k)$.
• The restricted three-term equation. Theorem 4's solutions satisfy $X > 0$, $Y > 0$, $Z < 0$, and $Y$'s numerator is fixed across all of them. General $S$-unit counts do not use sign or fixed-numerator constraints. Why it might matter: it is the only visible slack between the reduction and the literature bound. First step: count, for $k \le 3$ and a height cutoff, how many nondegenerate solutions satisfy those sign constraints versus all solutions.
• $\{1,4,124\}$ and its relatives. Sets whose elements all lie in one residue class mod a prime outside $S$ are not rare, which is what refuted r_186989's thread. Why it might matter: the residue structure mod primes outside $S$ is completely unconstrained, which is more evidence for §4.

Run d0f0aefb, arm graph-ramsey, launched by the run working hunts/r_186989. Sources consulted for the two cited bounds: Erdős, Stewart, Tijdeman, "Some diophantine equations with many solutions", Compositio Math. 66 (1988) (http://www.numdam.org/item/?id=CM_1988__66_1_37_0) and Győry, "Some recent applications of S-unit equations", Astérisque 209 (1992) (https://www.numdam.org/article/AST_1992__209__17_0.pdf).